Let `f(x)=(x(3^(x)-1))/(1-cosx)" for "x ne0`. Then value of `f(0)`, which make f(x) continuous at x = 0, is

Options

A. `log3`
B. `(1)/(2) log 3`
C. `(1)/(2log3)`
D. `2log3`

Answer

To find the value of \( f(0) \) that makes the function \( f(x) = \frac{x(3^x - 1)}{1 - \cos x} \) continuous at \( x = 0 \), we need to evaluate the limit of \( f(x) \) as \( x \) approaches 0 and set it equal to \( f(0) \).

### Step-by-Step Solution:

1. **Identify the limit to evaluate**:
We need to find:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{x(3^x - 1)}{1 - \cos x} \]

2. **Use L'Hôpital's Rule**:
Since both the numerator and denominator approach 0 as \( x \to 0 \), we can apply L'Hôpital's Rule. This states that if we have a limit of the form \( \frac{0}{0} \), we can take the derivative of the numerator and the denominator:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{d}{dx}[x(3^x - 1)] \bigg/ \frac{d}{dx}[1 - \cos x] \]

3. **Differentiate the numerator**:
Using the product rule:
\[ \frac{d}{dx}[x(3^x - 1)] = (3^x - 1) + x \cdot 3^x \ln(3) \]

4. **Differentiate the denominator**:
The derivative of \( 1 - \cos x \) is:
\[ \frac{d}{dx}[1 - \cos x] = \sin x \]

5. **Substituting back into the limit**:
Now we have:
\[ \lim_{x \to 0} \frac{(3^x - 1) + x \cdot 3^x \ln(3)}{\sin x} \]

6. **Evaluate the limit**:
As \( x \to 0 \):
- \( 3^x - 1 \to 0 \)
- \( \sin x \to 0 \)
- \( x \cdot 3^x \ln(3) \to 0 \)

We can apply L'Hôpital's Rule again since we still have the form \( \frac{0}{0} \).

7. **Differentiate again**:
Differentiate the numerator:
\[ \frac{d}{dx}[(3^x - 1) + x \cdot 3^x \ln(3)] = 3^x \ln(3) + (3^x \ln(3) + x \cdot 3^x (\ln(3))^2) \]
Simplifying gives:
\[ 3^x \ln(3)(1 + x \ln(3)) \]

The derivative of \( \sin x \) is \( \cos x \).

8. **Final limit**:
Now we evaluate:
\[ \lim_{x \to 0} \frac{3^x \ln(3)(1 + x \ln(3))}{\cos x} \]
As \( x \to 0 \):
- \( 3^x \to 1 \)
- \( \cos x \to 1 \)

Thus, the limit simplifies to:
\[ \ln(3) \]

9. **Set \( f(0) \)**:
To make \( f(x) \) continuous at \( x = 0 \), we set:
\[ f(0) = \ln(3) \]

### Conclusion:
The value of \( f(0) \) that makes \( f(x) \) continuous at \( x = 0 \) is:
\[ \boxed{\ln(3)} \]